Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Straight Line

A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10 m/s^2$. The fuel is finished in 1 minute. and it continuous to move up. What is the maximum height reached?

\[(a)\;18 \;km \quad (b)\;32 \;km \quad (c)\;12 \;km \quad(d)\;36\; km \]

Why did you use 0=(600)^2 - 2gh2

1 Answer

The distance travelled by the rocket in 1 minute (60 s) with $a=10 m/s^2$ upward is
$h_1= \large\frac{1}{2} $$\times 10 \times 60^2$
$\quad=18000 \;m$
$\quad =18 \;km$
Velocity acquired =>
$v=10 \times 60$
$\quad =600 m/s$
after 1 minute the rocket moves upward with velocity $600 m/s$ and down ward.
acceleration due to gravity till its velocity becomes zero
$0=(600)^2-2 gh_2$
$\quad= (600)^2 -2 \times 10 \times h_2$
$=>h_2=18000 \;m$
$=>18\; km$
Total distance $18+18=36\;km$
Hence d is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

Related questions